PHSC 121 - INTRODUCTION TO
ASTRONOMY - Kepler’s Laws of
Planetary Motion
Question Bank - Set 5
Liberty University
Question 1
Question
According to Kepler’s laws of planetary motion, the square of the orbital period
(T) of a planet is proportional to the cube of its average distance from the Sun
(r). Suppose a newly discovered planet has an orbital period of 12 years and an
average distance from the Sun of 5 AU (astronomical units). If another planet
has an average distance from the Sun of 15 AU, what is the approximate orbital
period of the second planet?
Solution
Step 1: Let’s denote the orbital period of the second planet as T2(in years) and
its average distance from the Sun as r2(in AU).
Step 2: According to Kepler’s third law, the relationship between the orbital
period and the average distance from the Sun is given by:
T2
1
r3
1
=T2
2
r3
2
where T1is the orbital period of the first planet (12 years) and r1is its average
distance from the Sun (5 AU).
Step 3: Substituting the given values into the equation, we have:
122
53=T2
2
153
Step 4: Simplifying the equation gives:
144
125 =T2
2
3375
Step 5: To find T2
2, we can cross multiply:
144 ×3375 = 125 ×T2
2
Step 6: Solving for T2
2gives:
T2
2=144 ×3375
125
Step 7: Calculate the value of T2:
T2=√144 ×3375
125
Step 8: Simplify the expression to find the approximate orbital period of the
second planet.
Question 2
Question
Kepler’s third law of planetary motion states that the square of the period of
revolution (T, in years) of a planet is directly proportional to the cube of the
semi-major axis of its orbit (a, in astronomical units). The proportionality can
be expressed as T2=k·a3, where kis a constant. Consider two planets, A and
B, with periods of revolution of 5 years and 10 years, respectively. If planet A
has a semi-major axis of orbit of 2 astronomical units, what is the semi-major
axis of orbit for planet B?
Solution
Step 1: Calculate the value of kfor each planet using Kepler’s third law. Step
2: Use the value of kand the given period of revolution for planet B to find the
semi-major axis of its orbit.
Step 1: For planet A:
T2
A=k·a3
A
52=k·23
25 = 8k
k=25
8= 3.125
For planet B:
T2
B=k·a3
B
102= 3.125 ·a3
B
100 = 3.125a3
B
a3
B=100
3.125 = 32
aB=3
√32 = 2
Therefore, the semi-major axis of orbit for planet B is 2 astronomical units.
2
Question 3
Question
Consider a planet with a semi-major axis of 2 AU orbiting a star with a mass
of 2×1030 kg. If the planet completes one full orbit in 2 years, determine its
orbital period, orbital speed, and gravitational force between the planet and the
star.
Solution
Step 1: Calculate the orbital period Tusing Kepler’s Third Law of Planetary
Motion:
T2=4π2
GM a3,
where T= orbital period, G= gravitational constant (6.67 ×10−11 m3/kg ·s2),
M= mass of the star (2×1030 kg), a= semi-major axis of the planet’s orbit
(2 AU).
Step 2: Substitute the given values into the equation to solve for T:
T2=4π2
(6.67 ×10−11)(2 ×1030)(2 AU)3
T2=4π2
1.334 ×1020 (8) ≈5.98 ×108
T≈√5.98 ×108≈24,464 days
Therefore, the orbital period of the planet is approximately 24,464 days.
Step 3: Calculate the orbital speed vof the planet using the formula:
v=2πa
T,
where v= orbital speed, a= semi-major axis of the planet’s orbit (2 AU), T=
orbital period (24,464 days).
Step 4: Substitute the values into the equation to find v:
v=2π(2)
24,464 ≈4π
24,464 ≈0.000257 AU/day
Therefore, the orbital speed of the planet is approximately 0.000257 AU/day.
Step 5: Calculate the gravitational force Fbetween the planet and the star
using Newton’s Law of Universal Gravitation:
F=GMm
r2,
where F= gravitational force, G= gravitational constant (6.67 ×10−11 m3/kg ·
s2), M= mass of the star (2×1030 kg), m= mass of the planet, r= distance
between the planet and the star.
3
Step 6: Assume the mass of the planet is negligible compared to the star,
then m≈0. The distance rcan be considered approximately equal to the
semi-major axis a.
Step 7: Substitute the values into the equation to find F:
F=(6.67 ×10−11)(2 ×1030)(0)
(2 AU)2= 0
Therefore, the gravitational force between the planet and the star is approx-
imately 0 N.
Question 4
Question
Given that a planet has an orbit with a semi-major axis of 2.5 AU, determine
the period of its orbit in years. Assume the planet’s orbit is nearly circular.
Solution
Step 1: Recall Kepler’s Third Law of Planetary Motion, which states that the
square of the period of a planet’s orbit (T) is proportional to the cube of the
semi-major axis of its orbit (a):
T2∝a3
Step 2: We can write the equation for Kepler’s Third Law as:
T2=k·a3
where kis the constant of proportionality.
Step 3: By rearranging the equation, we find:
T=√k·a3
1
Step 4: To determine the period of the planet’s orbit in years, we need to
find the value of the constant of proportionality (k). Since the planet’s orbit is
nearly circular, we can use Earth as a reference point. Earth has a semi-major
axis of 1 AU and a period of 1 year.
Step 5: Substituting T= 1 year and a= 1 AU into the equation gives:
1 = k·13
1 = k
Step 6: Now, we can plug k= 1 into the equation for the planet’s orbit:
T=√1·2.53
T=√15.625
T= 3.953 years
Therefore, the period of the planet’s orbit is approximately 3.953 years.
4
Question 5
Question
An exoplanet orbits a star with a semi-major axis of 0.25 astronomical units. If
the period of the exoplanet is 150 days, determine the mass of the star. (Hint:
Use Kepler’s third law).
Solution
Step 1: Recall Kepler’s third law which states that the square of the period (T)
of a planet is directly proportional to the cube of the semi-major axis (a) of its
orbit. Mathematically, this can be expressed as:
T2=k·a3
where kis a constant.
Step 2: We can rewrite the equation in terms of kby considering the data
given for the exoplanet:
T2
planet =k·a3
planet
(150 days)2=k·(0.25 AU)3
Step 3: Solve for the constant k:
(1502) = k·(0.253)
k=1502
0.253
Step 4: Now, using the mass of the star (Mstar) and the period of the star
(Tstar) in the modified Kepler’s third law, we have:
T2
star =k·a3
star
T2
star =1502
0.253·a3
star
Step 5: Given that astar = 0 for a star, the equation simplifies to:
T2
star =1502
0.253·0
Tstar = 0
Step 6: Since the period of a star is negligible, the mass of the star does not
affect the motion of the exoplanet.
Step 7: Therefore, the mass of the star is not required to determine the
exoplanet’s motion.
5
Question 6
Question
Suppose a spacecraft is in a circular orbit around a planet with a period of
20 hours. If the spacecraft’s altitude above the planet’s surface is 4,000 km,
determine the mass of the planet assuming the radius of the planet is 6,000 km.
Solution
Let’s assume that the planet can be approximated as a sphere. In a circular
orbit, the gravitational force provides the necessary centripetal force to keep the
spacecraft in its orbit. We can use Kepler’s third law: T2=4π2r3
GM , where: - T
is the period of the orbit, - ris the distance from the center of the planet to the
center of the spacecraft (altitude + radius of the planet), - Gis the gravitational
constant, - Mis the mass of the planet.
Step 1: Determine distance rfrom the center of the planet to the
center of the spacecraft
• Altitude above the planet’s surface: 4,000 km
• Radius of the planet: 6,000 km
Therefore, r=altitude +radius = 4,000 km + 6,000 km = 10,000 km.
Step 2: Calculate the period Tin seconds
• Given period T= 20 hours
Convert 20 hours to seconds: 20 ×60 ×60 = 72,000 s.
Step 3: Plug the known values into Kepler’s third law and solve
for the planet’s mass M
T2=4π2r3
GM
(72,000 s)2=4π2(10,000 km)3
G·M
5.184 ×109s2=4π2×1×109km3
G×M
Step 4: Solve for the mass Mof the planet
M=4π2×1×109km3
5.184 ×109s2×G
M=4π2×1×109km3
5.184 ×109s2×6.67430 ×10−11 m3kg−1s−2
M≈3.4×1022 kg
Therefore, the mass of the planet is approximately 3.4×1022 kg.
6
Question 7
Question
At a certain point in its orbit, a planet is 0.8 AU from the Sun and moving at a
speed of 28 km/s. Determine the shortest distance of the planet from the Sun
and the speed of the planet at this point in its orbit.
Given: a= 0.8AU, v= 28 km/s
Solution
Step 1: Determine the eccentricity of the planet’s orbit using Kepler’s Third
Law. T2
a3=4π2
GM
T2
(0.8)3=4π2
GM
T2
0.512 =4π2
GM
Solving for Tyields:
T=√0.512 ·4π2
GM
Step 2: Use the information provided to determine the instantaneous speed
and distance at the shortest distance from the Sun.
v2=GM (2
r−1
a)
Plug in the given values and solve for r:
282=GM (2
r−1
0.8)
Step 3: Determine the speed of the planet at its closest distance from the
Sun.
v′=√GM (2
r−1
a)
Substitute the value of robtained from step 2 and solve for v′.
Question 8
Question
According to Kepler’s third law of planetary motion, the square of the orbital
period of a planet is directly proportional to the cube of the semi-major axis of
7
its orbit. Suppose Planet X has an orbital period of 5 years and a semi-major
axis of 3 AU (astronomical units). Calculate the orbital period of Planet Y if it
has a semi-major axis of 8 AU.
Solution
Step 1: Write down Kepler’s third law in equation form:
T2∝a3
where Tis the orbital period and ais the semi-major axis.
Step 2: Use the given values for Planet X (T= 5 years, a= 3 AU) to set up
a proportion:
T2
1
a3
1
=T2
2
a3
2
Step 3: Substitute the values for Planet X into the proportion:
52
33=T2
2
83
Step 4: Simplify the equation:
25
27 =T2
2
512
Step 5: Cross multiply to solve for T2:
25 ×512 = 27 ×T2
2
T2
2=25 ×512
27
Step 6: Calculate T2:
T2
2=12800
27
T2=√12800
27
T2≈15.07 years
Therefore, the orbital period of Planet Y is approximately 15.07 years.
Question 9
Question
Given that the semi-major axis of an exoplanet’s orbit is 2.5 AU, determine the
period of its orbit in years.
8
Solution
Step 1: Recall Kepler’s Third Law of Planetary Motion:
T2=k×a3
where: T= period of orbit (in years) a= semi-major axis of the orbit (in AU)
k= proportionality constant
Step 2: Substitute the given semi-major axis (a= 2.5AU) into the equation.
T2=k×(2.5)3
Step 3: Since kis a constant, we can rewrite the equation as:
T2=k×15.625
Step 4: To solve for T, we need to find the value of the constant k. This
can be done by using data from a known system, such as the Earth’s orbit with
a= 1 AU and T= 1 year.
1 = k×13
k= 1
Step 5: Substitute k= 1 back into the equation:
T2= 15.625
Step 6: Take the square root of both sides to solve for T:
T=√15.625 = 3.944
Therefore, the period of the exoplanet’s orbit is approximately 3.944 years.
Question 10
Question
An exoplanet orbits its star in a period of 30 days at an average distance of 0.1
AU. If the mass of the star is known to be 2×1030 kg, determine the mass of
the exoplanet.
Solution
Step 1: We can use Kepler’s third law of planetary motion to determine the
mass of the exoplanet. Kepler’s third law states that the square of the orbital
period of a planet is directly proportional to the cube of the semi-major axis of
its orbit. Mathematically, this can be expressed as:
T2
1
T2
2
=a3
1
a3
2
9
where T1and T2are the orbital periods of the two planets, and a1and a2are
their semi-major axes.
Step 2: Given that the exoplanet’s orbital period is 30 days and its average
distance is 0.1 AU, and that the star’s mass is 2×1030 kg, we can set up the
equation:
302
T2
2
=0.13
a3
2
Step 3: To solve for the mass of the exoplanet, we need to find its semi-
major axis. Since the exoplanet orbits the star, the semi-major axis is the
average distance between the exoplanet and the star:
semi-major axis =average distance
Step 4: Substituting the given values into the equation, we get:
302
T2
2
=0.13
0.13
Step 5: Simplifying the equation further, we find:
900
T2
2
= 1
Step 6: Solving for the orbital period of the exoplanet, we get:
T2
2= 900 ⇒T2= 30 days
Step 7: Now that we have the orbital period of the exoplanet, we can use
Kepler’s third law to find its mass. Kepler’s third law in this case simplifies to:
M1=(T1
T2)2
×M2
Step 8: Substituting the known values into the equation, we can now solve
for the mass of the exoplanet:
M1=(30
30)2
×2×1030 = 2 ×1030 kg
Step 9: Therefore, the mass of the exoplanet is 2×1030 kg.
Question 11
Question
Consider a planet in a circular orbit around a star. The planet takes 2 years to
complete one orbit. If the distance between the planet and the star is 2×108
km, calculate the mass of the star in terms of the mass of the planet.
10
Solution
Step 1: We can start by using Kepler’s third law of planetary motion which
states that the square of the period of revolution of a planet is directly propor-
tional to the cube of its mean distance from the Sun. Mathematically, this can
be expressed as:
T2
1
R3
1
=T2
2
R3
2
where T1and R1are the period and distance of the unknown mass star, and T2
and R2are the given period and distance of the planet.
Step 2: We are given that the period T2of the planet is 2 years, so T2= 2
years. The distance R2between the planet and the star is 2×108km or
R2= 2 ×108km.
Step 3: The period T1of the star is unknown and the distance R1is also
unknown. Let the mass of the planet be represented by Mpand the mass of the
star be represented by Ms.
Step 4: We know that the mass of the star Msis directly proportional to
the cube of the distance R1and the square of the period T1. We can rewrite
the formula in terms of mass as:
Ms
Mp
=(R1
R2)3(T1
T2)2
Step 5: Substituting the given values T2= 2 years and R2= 2 ×108km into
the equation above, we get:
Ms
Mp
=(R1
2×108)3(T1
2)2
Step 6: Simplifying, we find that:
Ms
Mp
=(R1
2×108)3(T1
2)2
= 1
Step 7: Therefore, the mass of the star, in terms of the mass of the planet,
is equal to the ratio of their masses which is 1. Hence, the mass of the star is
equal to the mass of the planet.
Question 12
Question
Two planets, Planet A and Planet B, orbit a star. Planet A has a semi-major
axis of 3 AU and an orbital period of 4 years. Planet B has a semi-major axis
of 4 AU and an orbital period of 6 years. Determine which planet has a greater
orbital speed and by how much.
11
Solution
Step 1: Use Kepler’s Third Law to find the orbital speed of each planet. Ke-
pler’s Third Law states that the square of the orbital period (T) of a planet is
proportional to the cube of the semi-major axis (a) of its orbit. Mathematically,
this can be expressed as:
T2
A
a3
A
=T2
B
a3
B
where TAand aAare the orbital period and semi-major axis of Planet A, while
TBand aBare the orbital period and semi-major axis of Planet B.
Step 2: Using the given values for Planet A: TA= 4 years, aA= 3 AU.
Substitute these values into the equation:
42
33=16
27 =T2
B
a3
B
Step 3: Using the given values for Planet B: TB= 6 years, aB= 4 AU. Solve
for T2
B:
T2
B=16
27 ×a3
B=16
27 ×43=16
27 ×64
T2
B=16 ×64
27
T2
B=1024
27
Step 4: Now, find the orbital speed vof each planet using the formula:
v=2πa
T
where vis the orbital speed, ais the semi-major axis, and Tis the orbital period.
Step 5: For Planet A: a= 3 AU, T= 4 years.
vA=2π×3
4=6π
4
vA=3π
2
Step 6: For Planet B: a= 4 AU, T= 6 years.
vB=2π×4
6=8π
6
vB=4π
3
Step 7: Compare the orbital speeds of the two planets: vA=3π
2and vB=
4π
3. Thus, the greater orbital speed is vB=4π
3AU/year.
Therefore, Planet B has a greater orbital speed than Planet A by π
6AU/year.
12
Question 13
Question
According to Kepler’s third law of planetary motion, the square of the orbital
period of a planet is proportional to the cube of its semi-major axis. Given that
the orbital period of Earth is 365.25 days and its average distance from the Sun
is approximately 1 astronomical unit (AU), determine the orbital period of a
hypothetical planet with an average distance from the Sun of 3 AU.
Solution
Step 1: Write down Kepler’s third law in mathematical form:
Kepler’s third law states that the ratio of the squares of the periods of
any two planets is equal to the ratio of the cubes of their semi-major axes.
Mathematically, this can be expressed as:
(T1
T2)2
=(a1
a2)3
where T1and T2are the periods of the two planets, and a1and a2are their
semi-major axes.
Step 2: Given that the orbital period of Earth (T1) is 365.25 days and its
average distance from the Sun (a1) is 1 AU, we are asked to find the orbital
period of a planet with an average distance from the Sun of 3 AU (a2).
Step 3: Substitute the known values into Kepler’s third law equation:
(365.25
T2)2
=(1
3)3
Step 4: Solve for T2:
(365.25
T2)2
=(1
27)
365.25
T2
=√1
27
365.25
T2
=1
3
3×365.25 = T2
T2= 1095.75 days
Therefore, the orbital period of the hypothetical planet with an average
distance from the Sun of 3 AU is approximately 1095.75 days.
13
Question 14
Question
Kepler’s second law states that a planet moves fastest when it is closest to the
Sun (at perihelion) and slowest when it is farthest from the Sun (at aphelion).
Suppose a planet has a perihelion distance of 0.5 AU and an aphelion distance
of 1.5 AU. Calculate the average speed of the planet in its elliptical orbit in
terms of its speed at perihelion.
Solution
Step 1: Let’s denote the planet’s speed at perihelion as vp. The planet covers
equal areas in equal times, so we have:
Area swept at perihelion =Area swept at aphelion
Step 2: The area swept by the planet at perihelion can be calculated using the
formula for the area of an ellipse:
Area =π×semi-major axis ×semi-minor axis
Thus, at perihelion:
π×0.5×b=π×0.5vp
Step 3: Similarly, at aphelion:
π×1.5×b=π×1.5vp
⇒1.5×b= 0.5×b
⇒b=1
3b
Step 4: The average speed of the planet in its orbit can be calculated using the
formula:
Average speed =total distance
total time
Step 5: The total distance covered by the planet in one complete orbit is the
circumference of an ellipse, given by the formula:
2π×√a2+b2
2
where ais the semi-major axis and bis the semi-minor axis. Step 6: The
total time to complete one elliptical orbit is the same whether the planet is at
perihelion or aphelion. The time taken to travel from perihelion to aphelion
is equal to the time taken to travel from aphelion back to perihelion. Step 7:
Using the equations from Steps 2 and 3 along with the information from Step 5,
we can calculate the average speed of the planet in its elliptical orbit in terms
of its speed at perihelion.
14
Question 15
Question
The asteroid Eris has an average orbital radius from the Sun of approximately
10 billion kilometers. If the orbital period of Eris is about 560 years, what is
the mass of the Sun? (Assume Eris follows a circular orbit.)
Solution
Step 1: We can use Kepler’s Third Law of Planetary Motion to relate the orbital
period of a planet to its average orbital radius. Kepler’s Third Law states that
the square of the orbital period of a planet is directly proportional to the cube
of the average orbital radius. Mathematically, this can be written as:
T2=k×R3
where Tis the orbital period, Ris the average orbital radius, and kis a constant
of proportionality. Step 2: We can rearrange the equation to solve for the
constant k:
k=T2
R3
Step 3: Since the asteroid Eris has an average orbital radius of 10 billion kilo-
meters (or 10 ×109km) and an orbital period of 560 years, we can substitute
these values into the equation to find the constant k:
k=(560 years)2
(10 ×109km)3
Step 4: Calculate the value of k:
k=313600 years2
109×109×109km3=313600 years2
1027 km3
k= 3.136 ×10−14 years2/km3
Step 5: Now that we know the constant k, we can use it to find the mass of
the Sun using the average orbital radius of the Earth (1.496 ×108km) and the
orbital period of Earth (1 year):
mSun =4π2
k
R3
T2
mSun =4π2
3.136 ×10−14
(10 ×109)3
(560)2
Step 6: Calculate the mass of the Sun:
mSun =4π2
3.136 ×10−14
1027
313600
15
mSun =4×(π)2×1027
3.136 ×313600
mSun ≈2.03 ×1030 kg
Therefore, the mass of the Sun is approximately 2.03 ×1030 kg.
Question 16
Question
Kepler’s Third Law of Planetary Motion states that the square of the orbital
period of a planet is proportional to the cube of its semi-major axis. Suppose a
planet has an orbital period of 10 years and a semi-major axis of 3 astronomical
units (AU). Calculate the orbital period of another planet with a semi-major
axis of 6 AU.
Solution
To apply Kepler’s Third Law of Planetary Motion, we can use the formula:
T2
1
a3
1
=T2
2
a3
2
where: - T1and a1are the orbital period and semi-major axis of the first planet,
respectively, - T2and a2are the orbital period and semi-major axis of the second
planet, respectively.
Step 1: Substitute the values given for the first planet: T1= 10 years and
a1= 3 AU, into the formula.
102
33=T2
2
63
Step 2: Simplify the equation.
100
27 =T2
2
216
Step 3: Cross multiply to find T2
2.
T2
2=100
27 ×216
T2
2=100 ×216
27
T2
2=21600
27
T2
2= 800
16
Step 4: Take the square root of both sides to find T2.
T2=√800
T2= 28.28 years
Therefore, the orbital period of the other planet with a semi-major axis of
6 AU is approximately 28.28 years.
Question 17
Question
Explain Kepler’s Third Law of Planetary Motion and how it relates to the period
and distance of a planet from the Sun.
Solution
Kepler’s Third Law of Planetary Motion states that the square of the period
of revolution of a planet around the Sun is proportional to the cube of the
semi-major axis of its orbit. Mathematically, this can be expressed as:
T2=k·a3
where: - Tis the period of revolution of the planet, - ais the semi-major axis
of the planet’s orbit, - kis a constant of proportionality.
Step 1: Let’s denote the period of revolution of Planet A as TA, the semi-
major axis of its orbit as aA, and the constant of proportionality for Planet A
as kA.
Step 2: Similarly, denote the period of revolution of Planet B as TB, the
semi-major axis of its orbit as aB, and the constant of proportionality for Planet
B as kB.
Step 3: Using Kepler’s Third Law, we have the following equations for
Planets A and B:
T2
A=kA·a3
A
T2
B=kB·a3
B
Step 4: Now, let’s compare the two planets. Since both planets revolve
around the same Sun, we can set kA=kB=kfor simplicity.
Step 5: By setting the two equations equal to each other, we get:
T2
A=T2
B
k·a3
A=k·a3
B
Step 6: Simplifying the above equation, we find that:
a3
A=a3
B
17
Step 7: Taking the cube root of both sides, we get:
aA=aB
Step 8: Therefore, the semi-major axes of the orbits of both planets are
equal. This means that the distance of a planet from the Sun is directly related
to its period of revolution.
Question 18
Question
Kepler’s third law of planetary motion states that the square of the period of
revolution of a planet is proportional to the cube of its average distance from
the Sun. Suppose a newly discovered planet, Planet X, has an average distance
from the Sun that is twice that of Earth. If Earth’s period of revolution around
the Sun is 1 year, what is Planet X’s period of revolution in terms of Earth
years?
Solution
Step 1: Let the average distance of Earth from the Sun be represented as rE,
and the period of revolution of Earth be denoted as TE. Given that the average
distance of Planet X from the Sun is twice that of Earth’s, we have the average
distance of Planet X, rX= 2rE.
Step 2: According to Kepler’s third law, the square of the period of revolution
of a planet is proportional to the cube of its average distance from the Sun.
Mathematically, this relationship is represented as:
(TX
TE)2
=(rX
rE)3
Step 3: Substituting the values for rXand rE, we have:
(TX
TE)2
=(2rE
rE)3
= 23
Step 4: Simplifying the equation, we get:
(TX
TE)2
= 8
Step 5: Taking the square root of both sides, we find:
TX
TE
=√8 = 2√2
Step 6: Therefore, the period of revolution of Planet X in terms of Earth
years is TX= 2√2×TE= 2√2×1 = 2√2years. Thus, the period of revolution
of Planet X is approximately 2.83 Earth years.
18
Question 19
Question
An asteroid orbits the Sun with a semi-major axis of 3.2 AU. If the time taken
for the asteroid to complete one orbit is 5.6 years, determine the mass of the
Sun in terms of the mass of the Earth. Assume the asteroid has a negligible
mass compared to the Sun.
Solution
Step 1: Find the Sun’s mass in terms of the asteroid’s orbital period.
T2=4π2a3
GM
where: T= 5.6years (orbital period), a= 3.2AU (semi-major axis), G=
universal gravitational constant, and M= mass of the Sun.
Step 2: Substitute the given values into the equation.
(5.6)2=4π2(3.2)3
G·M
Step 3: Solve for M.
M=4π2(3.2)3
G·(5.6)2
Step 4: Approximate the value of Min terms of Earth’s mass.
M=4π2(3.2)3
(6.674 ×10−11)·(5.6)2
M≈3.68 ×105M⊕
Therefore, the mass of the Sun in terms of the mass of the Earth is approx-
imately 3.68 ×105times the mass of the Earth.
Question 20
Question
Consider a hypothetical planet with an orbital period of 200 days. If the semi-
major axis of the planet’s orbit is 2 AU (astronomical units), what is the mass
of the central star in solar masses? (Assume the star is much more massive than
the planet.)
19
Solution
Step 1: Recall Kepler’s third law, which relates the orbital period of a planet
(T), the semi-major axis of its orbit (a), and the mass of the central star (M)
using the equation:
T2=4π2
GM a3
where Gis the gravitational constant.
Step 2: We are given that T= 200 days (1day = 24 ×60 ×60 seconds) and
a= 2 AU. We know that 1AU = 1.496 ×108km. Converting afrom AU to
km:
a= 2 AU ×1.496 ×108km/AU = 2.992 ×108km
Step 3: We now have all the values needed to solve for the mass of the
central star. Substituting T,a, and the known constants into Kepler’s third law
equation:
(200 ×24 ×60 ×60)2=4π2
6.67 ×10−11M(2.992 ×108)3
Step 4: Simplify the equation by squaring 200 ×24 ×60 ×60 and cubing
2.992 ×108. Then solve for M:
3.456 ×109=4π2
6.67 ×10−11M2.160 ×1024
Step 5: Rearrange the equation to solve for M:
M=4π2
6.67 ×10−11 ×3.456 ×109(2.160 ×1024)
Step 6: Calculate the value of Musing a calculator to find the mass of the
central star in solar masses.
Question 21
Question
Explain Kepler’s Third Law of Planetary Motion and calculate the orbital period
of a hypothetical planet orbiting a star with a semimajor axis of 3 AU.
Solution
Kepler’s Third Law of Planetary Motion states that the square of the orbital
period (T) of a planet is directly proportional to the cube of the semimajor axis
(a) of its orbit. Mathematically, it can be expressed as:
T2=k×a3
20
where kis a constant that depends on the system of measurement used.
Step 1: Given that the semimajor axis a= 3 AU, we can find the orbital
period Tin years.
Step 2: Since we are dealing with the square of the orbital period and the
cube of the semimajor axis, we need to set up a proportion to solve for T.
T2=k×a3
T2=k×33
T2= 27k
Step 3: Now, if we have information about the Earth’s orbital period and
semimajor axis, we can use those values to find the constant k. Let’s say TEarth =
1year and aEarth = 1 AU. Plugging these values into the equation:
12= 27k
1 = 27k
k=1
27
Step 4: Substituting the value of kback into the equation, we can solve for
the orbital period Tof the hypothetical planet.
T2=1
27 ×33
T2=1
27 ×27
T2= 1
T= 1 year
Therefore, the orbital period of the hypothetical planet orbiting a star with
a semimajor axis of 3 AU would be 1 year.
Question 22
Question
According to Kepler’s Laws of Planetary Motion, a planet moves fastest when
it is closest to the Sun. Suppose a planet takes 200 days to complete one orbit
around the Sun. If the planet is closest to the Sun at a distance of 0.3 AU and
farthest from the Sun at a distance of 0.7 AU, determine the average speed and
the maximum speed of the planet in its orbit (in AU/day).
21
Solution
Step 1: To find the average speed of the planet, we will use the formula for
average speed:
Average Speed =Total Distance
Total Time
Step 2: The total distance the planet travels during one orbit can be cal-
culated using the sum of the distances when the planet is closest and farthest
from the Sun:
Total Distance =Distance when closest +Distance when farthest
Step 3: Given that the planet is closest to the Sun at 0.3 AU and farthest
at 0.7 AU, we have:
Total Distance = 0.3AU + 0.7AU = 1.0AU
Step 4: Given that the planet takes 200 days to complete one orbit, the total
time is 200 days:
Total Time = 200 days
Step 5: Now we can calculate the average speed of the planet:
Average Speed =1.0AU
200 days = 0.005 AU/day
Step 6: To find the maximum speed of the planet, we note that the planet
moves fastest when closest to the Sun. Thus, the maximum speed of the planet
is the speed when it is at 0.3 AU.
Step 7: Since the planet takes 200 days to complete one orbit, at 0.3 AU,
the planet travels a distance of 0.3 AU in half of an orbit (i.e., 100 days).
Max Speed =0.3AU
100 days = 0.003 AU/day
Therefore, the average speed of the planet is 0.005 AU/day, and the maxi-
mum speed of the planet in its orbit is 0.003 AU/day.
Question 23
Question
A planet orbits a star in an elliptical orbit where the distance between the
planet and the star at closest approach is 28 million kilometers and the distance
at farthest approach is 100 million kilometers. If the planet takes 200 days to
complete one orbit around the star, what is the velocity of the planet when it
is at its closest approach to the star?
22
Solution
Step 1: Find the semi-major axis of the planet’s orbit. This can be calculated as
the average of the closest and farthest distances from the star. Let rclosest = 28
million km and rfarthest = 100 million km. The semi-major axis is given by:
a=rclosest +rfarthest
2
a=28 + 100
2= 64 million km = 64 ×106km
Step 2: Calculate the eccentricity of the orbit. The eccentricity of the ellipse
can be calculated as:
e=rfarthest −rclosest
rfarthest +rclosest
Substitute the values:
e=100 −28
100 + 28 =72
128 = 0.5625
Step 3: Use Kepler’s Second Law to find the velocity of the planet at its
closest approach. The law states that the line joining the planet and the sun
sweeps out equal areas in equal intervals of time. At closest approach, the planet
moves fastest. The velocity of the planet at any position can be found using the
equation:
v=√GM(1 + e)
a(1 −e)
where vis the velocity, Gis the gravitational constant, Mis the mass of the
star, ais the semi-major axis, and eis the eccentricity of the orbit. At the
closest approach, r=rclosest:
vclosest =√GM(1 + e)
a(1 −e)
Substitute the given values and constants into the equation:
vclosest =√6.674 ×10−11 ×M(1 + 0.5625)
64 ×106×(1 −0.5625)
vclosest =√6.674 ×10−11 ×M×1.5625
64 ×106×0.4375
vclosest =√10.41875 ×10−11 ×M
28 ×106
vclosest =√0.37245 ×10−5×M
Therefore, the velocity of the planet when it is at its closest approach to the
star is √0.37245 ×10−5×M.
23
Question 24
Question
The orbit of a certain planet is known to be an ellipse with semi-major axis of
2.5×106km. If the planet has an eccentricity of 0.4, determine the distance
between the planet and the Sun at its closest approach.
Solution
Step 1: Recall Kepler’s Second Law, which states that a planet will sweep out
equal areas in equal times as it travels along its orbit. When a planet is at its
closest approach to the Sun (called perihelion), it moves fastest. This means
the line connecting the Sun and the planet sweeps out an equal area in a given
time compared to other points in the orbit.
Step 2: The distance between the planet and the Sun at closest approach
is equal to the semi-minor axis of the elliptical orbit. The semi-major axis and
the eccentricity of the orbit can be used to calculate the semi-minor axis using
the formula:
b=a√1−e2
where ais the semi-major axis, eis the eccentricity, and bis the semi-minor
axis.
Step 3: Plug in the given values:
b= 2.5×106km ×√1−0.42
b= 2.5×106km ×√1−0.16
b= 2.5×106km ×√0.84
b= 2.5×106km ×0.916515
b≈2.2912875 ×106km
Step 4: Therefore, the distance between the planet and the Sun at its closest
approach is approximately 2.291 ×106km.
Question 25
Question
Suppose a planet has an elliptical orbit around the Sun such that its eccentricity
is 0.4. If the distance between the planet and the Sun at its closest approach
(perihelion) is 0.3 astronomical units (AU), determine the distance between the
planet and the Sun at its farthest point (aphelion).
24
Solution
Step 1: Recall that the eccentricity of an elliptical orbit is defined as the ratio
of the distance between the foci of the ellipse to the length of the major axis.
Mathematically, the eccentricity (e) can be expressed as:
e=c
a
where cis half the distance between the foci of the ellipse, and ais half the
length of the major axis.
Step 2: We are given that the eccentricity of the planet’s orbit is 0.4, and
the distance between the planet and the Sun at perihelion is 0.3 AU. This
information allows us to set up an equation to solve for the distance between
the planet and the Sun at aphelion.
Step 3: Let the distance between the planet and the Sun at aphelion be x
AU. From the definition of eccentricity, we have:
0.4 = c
a=x−0.3
x+0.3
2
Step 4: Simplifying the equation, we get:
0.4 = 2(x−0.3)
x+ 0.3
Step 5: Cross multiplying gives:
0.4(x+ 0.3) = 2(x−0.3)
Step 6: Expanding both sides of the equation yields:
0.4x+ 0.12 = 2x−0.6
Step 7: Rearranging the terms gives:
0.12 + 0.6 = 2x−0.4x
0.72 = 1.6x
Step 8: Finally, solving for x, we find:
x=0.72
1.6≈0.45 AU
Step 9: Therefore, the distance between the planet and the Sun at its farthest
point (aphelion) is approximately 0.45 AU.
25
Step 5: To find T2
2, we can cross multiply:
144 ×3375 = 125 ×T2
2
Step 6: Solving for T2
2gives:
T2
2=144 ×3375
125
Step 7: Calculate the value of T2:
T2=√144 ×3375
125
Step 8: Simplify the expression to find the approximate orbital period of the
second planet.
Question 2
Question
Kepler’s third law of planetary motion states that the square of the period of
revolution (T, in years) of a planet is directly proportional to the cube of the
semi-major axis of its orbit (a, in astronomical units). The proportionality can
be expressed as T2=k·a3, where kis a constant. Consider two planets, A and
B, with periods of revolution of 5 years and 10 years, respectively. If planet A
has a semi-major axis of orbit of 2 astronomical units, what is the semi-major
axis of orbit for planet B?
Solution
Step 1: Calculate the value of kfor each planet using Kepler’s third law. Step
2: Use the value of kand the given period of revolution for planet B to find the
semi-major axis of its orbit.
Step 1: For planet A:
T2
A=k·a3
A
52=k·23
25 = 8k
k=25
8= 3.125
For planet B:
T2
B=k·a3
B
102= 3.125 ·a3
B
100 = 3.125a3
B
a3
B=100
3.125 = 32
aB=3
√32 = 2
Therefore, the semi-major axis of orbit for planet B is 2 astronomical units.
2
Question 3
Question
Consider a planet with a semi-major axis of 2 AU orbiting a star with a mass
of 2×1030 kg. If the planet completes one full orbit in 2 years, determine its
orbital period, orbital speed, and gravitational force between the planet and the
star.
Solution
Step 1: Calculate the orbital period Tusing Kepler’s Third Law of Planetary
Motion:
T2=4π2
GM a3,
where T= orbital period, G= gravitational constant (6.67 ×10−11 m3/kg ·s2),
M= mass of the star (2×1030 kg), a= semi-major axis of the planet’s orbit
(2 AU).
Step 2: Substitute the given values into the equation to solve for T:
T2=4π2
(6.67 ×10−11)(2 ×1030)(2 AU)3
T2=4π2
1.334 ×1020 (8) ≈5.98 ×108
T≈√5.98 ×108≈24,464 days
Therefore, the orbital period of the planet is approximately 24,464 days.
Step 3: Calculate the orbital speed vof the planet using the formula:
v=2πa
T,
where v= orbital speed, a= semi-major axis of the planet’s orbit (2 AU), T=
orbital period (24,464 days).
Step 4: Substitute the values into the equation to find v:
v=2π(2)
24,464 ≈4π
24,464 ≈0.000257 AU/day
Therefore, the orbital speed of the planet is approximately 0.000257 AU/day.
Step 5: Calculate the gravitational force Fbetween the planet and the star
using Newton’s Law of Universal Gravitation:
F=GMm
r2,
where F= gravitational force, G= gravitational constant (6.67 ×10−11 m3/kg ·
s2), M= mass of the star (2×1030 kg), m= mass of the planet, r= distance
between the planet and the star.
3
Step 6: Assume the mass of the planet is negligible compared to the star,
then m≈0. The distance rcan be considered approximately equal to the
semi-major axis a.
Step 7: Substitute the values into the equation to find F:
F=(6.67 ×10−11)(2 ×1030)(0)
(2 AU)2= 0
Therefore, the gravitational force between the planet and the star is approx-
imately 0 N.
Question 4
Question
Given that a planet has an orbit with a semi-major axis of 2.5 AU, determine
the period of its orbit in years. Assume the planet’s orbit is nearly circular.
Solution
Step 1: Recall Kepler’s Third Law of Planetary Motion, which states that the
square of the period of a planet’s orbit (T) is proportional to the cube of the
semi-major axis of its orbit (a):
T2∝a3
Step 2: We can write the equation for Kepler’s Third Law as:
T2=k·a3
where kis the constant of proportionality.
Step 3: By rearranging the equation, we find:
T=√k·a3
1
Step 4: To determine the period of the planet’s orbit in years, we need to
find the value of the constant of proportionality (k). Since the planet’s orbit is
nearly circular, we can use Earth as a reference point. Earth has a semi-major
axis of 1 AU and a period of 1 year.
Step 5: Substituting T= 1 year and a= 1 AU into the equation gives:
1 = k·13
1 = k
Step 6: Now, we can plug k= 1 into the equation for the planet’s orbit:
T=√1·2.53
T=√15.625
T= 3.953 years
Therefore, the period of the planet’s orbit is approximately 3.953 years.
4
Question 5
Question
An exoplanet orbits a star with a semi-major axis of 0.25 astronomical units. If
the period of the exoplanet is 150 days, determine the mass of the star. (Hint:
Use Kepler’s third law).
Solution
Step 1: Recall Kepler’s third law which states that the square of the period (T)
of a planet is directly proportional to the cube of the semi-major axis (a) of its
orbit. Mathematically, this can be expressed as:
T2=k·a3
where kis a constant.
Step 2: We can rewrite the equation in terms of kby considering the data
given for the exoplanet:
T2
planet =k·a3
planet
(150 days)2=k·(0.25 AU)3
Step 3: Solve for the constant k:
(1502) = k·(0.253)
k=1502
0.253
Step 4: Now, using the mass of the star (Mstar) and the period of the star
(Tstar) in the modified Kepler’s third law, we have:
T2
star =k·a3
star
T2
star =1502
0.253·a3
star
Step 5: Given that astar = 0 for a star, the equation simplifies to:
T2
star =1502
0.253·0
Tstar = 0
Step 6: Since the period of a star is negligible, the mass of the star does not
affect the motion of the exoplanet.
Step 7: Therefore, the mass of the star is not required to determine the
exoplanet’s motion.
5
Question 6
Question
Suppose a spacecraft is in a circular orbit around a planet with a period of
20 hours. If the spacecraft’s altitude above the planet’s surface is 4,000 km,
determine the mass of the planet assuming the radius of the planet is 6,000 km.
Solution
Let’s assume that the planet can be approximated as a sphere. In a circular
orbit, the gravitational force provides the necessary centripetal force to keep the
spacecraft in its orbit. We can use Kepler’s third law: T2=4π2r3
GM , where: - T
is the period of the orbit, - ris the distance from the center of the planet to the
center of the spacecraft (altitude + radius of the planet), - Gis the gravitational
constant, - Mis the mass of the planet.
Step 1: Determine distance rfrom the center of the planet to the
center of the spacecraft
• Altitude above the planet’s surface: 4,000 km
• Radius of the planet: 6,000 km
Therefore, r=altitude +radius = 4,000 km + 6,000 km = 10,000 km.
Step 2: Calculate the period Tin seconds
• Given period T= 20 hours
Convert 20 hours to seconds: 20 ×60 ×60 = 72,000 s.
Step 3: Plug the known values into Kepler’s third law and solve
for the planet’s mass M
T2=4π2r3
GM
(72,000 s)2=4π2(10,000 km)3
G·M
5.184 ×109s2=4π2×1×109km3
G×M
Step 4: Solve for the mass Mof the planet
M=4π2×1×109km3
5.184 ×109s2×G
M=4π2×1×109km3
5.184 ×109s2×6.67430 ×10−11 m3kg−1s−2
M≈3.4×1022 kg
Therefore, the mass of the planet is approximately 3.4×1022 kg.
6
Question 7
Question
At a certain point in its orbit, a planet is 0.8 AU from the Sun and moving at a
speed of 28 km/s. Determine the shortest distance of the planet from the Sun
and the speed of the planet at this point in its orbit.
Given: a= 0.8AU, v= 28 km/s
Solution
Step 1: Determine the eccentricity of the planet’s orbit using Kepler’s Third
Law. T2
a3=4π2
GM
T2
(0.8)3=4π2
GM
T2
0.512 =4π2
GM
Solving for Tyields:
T=√0.512 ·4π2
GM
Step 2: Use the information provided to determine the instantaneous speed
and distance at the shortest distance from the Sun.
v2=GM (2
r−1
a)
Plug in the given values and solve for r:
282=GM (2
r−1
0.8)
Step 3: Determine the speed of the planet at its closest distance from the
Sun.
v′=√GM (2
r−1
a)
Substitute the value of robtained from step 2 and solve for v′.
Question 8
Question
According to Kepler’s third law of planetary motion, the square of the orbital
period of a planet is directly proportional to the cube of the semi-major axis of
7
its orbit. Suppose Planet X has an orbital period of 5 years and a semi-major
axis of 3 AU (astronomical units). Calculate the orbital period of Planet Y if it
has a semi-major axis of 8 AU.
Solution
Step 1: Write down Kepler’s third law in equation form:
T2∝a3
where Tis the orbital period and ais the semi-major axis.
Step 2: Use the given values for Planet X (T= 5 years, a= 3 AU) to set up
a proportion:
T2
1
a3
1
=T2
2
a3
2
Step 3: Substitute the values for Planet X into the proportion:
52
33=T2
2
83
Step 4: Simplify the equation:
25
27 =T2
2
512
Step 5: Cross multiply to solve for T2:
25 ×512 = 27 ×T2
2
T2
2=25 ×512
27
Step 6: Calculate T2:
T2
2=12800
27
T2=√12800
27
T2≈15.07 years
Therefore, the orbital period of Planet Y is approximately 15.07 years.
Question 9
Question
Given that the semi-major axis of an exoplanet’s orbit is 2.5 AU, determine the
period of its orbit in years.
8
Solution
Step 1: Recall Kepler’s Third Law of Planetary Motion:
T2=k×a3
where: T= period of orbit (in years) a= semi-major axis of the orbit (in AU)
k= proportionality constant
Step 2: Substitute the given semi-major axis (a= 2.5AU) into the equation.
T2=k×(2.5)3
Step 3: Since kis a constant, we can rewrite the equation as:
T2=k×15.625
Step 4: To solve for T, we need to find the value of the constant k. This
can be done by using data from a known system, such as the Earth’s orbit with
a= 1 AU and T= 1 year.
1 = k×13
k= 1
Step 5: Substitute k= 1 back into the equation:
T2= 15.625
Step 6: Take the square root of both sides to solve for T:
T=√15.625 = 3.944
Therefore, the period of the exoplanet’s orbit is approximately 3.944 years.
Question 10
Question
An exoplanet orbits its star in a period of 30 days at an average distance of 0.1
AU. If the mass of the star is known to be 2×1030 kg, determine the mass of
the exoplanet.
Solution
Step 1: We can use Kepler’s third law of planetary motion to determine the
mass of the exoplanet. Kepler’s third law states that the square of the orbital
period of a planet is directly proportional to the cube of the semi-major axis of
its orbit. Mathematically, this can be expressed as:
T2
1
T2
2
=a3
1
a3
2
9
where T1and T2are the orbital periods of the two planets, and a1and a2are
their semi-major axes.
Step 2: Given that the exoplanet’s orbital period is 30 days and its average
distance is 0.1 AU, and that the star’s mass is 2×1030 kg, we can set up the
equation:
302
T2
2
=0.13
a3
2
Step 3: To solve for the mass of the exoplanet, we need to find its semi-
major axis. Since the exoplanet orbits the star, the semi-major axis is the
average distance between the exoplanet and the star:
semi-major axis =average distance
Step 4: Substituting the given values into the equation, we get:
302
T2
2
=0.13
0.13
Step 5: Simplifying the equation further, we find:
900
T2
2
= 1
Step 6: Solving for the orbital period of the exoplanet, we get:
T2
2= 900 ⇒T2= 30 days
Step 7: Now that we have the orbital period of the exoplanet, we can use
Kepler’s third law to find its mass. Kepler’s third law in this case simplifies to:
M1=(T1
T2)2
×M2
Step 8: Substituting the known values into the equation, we can now solve
for the mass of the exoplanet:
M1=(30
30)2
×2×1030 = 2 ×1030 kg
Step 9: Therefore, the mass of the exoplanet is 2×1030 kg.
Question 11
Question
Consider a planet in a circular orbit around a star. The planet takes 2 years to
complete one orbit. If the distance between the planet and the star is 2×108
km, calculate the mass of the star in terms of the mass of the planet.
10
Solution
Step 1: We can start by using Kepler’s third law of planetary motion which
states that the square of the period of revolution of a planet is directly propor-
tional to the cube of its mean distance from the Sun. Mathematically, this can
be expressed as:
T2
1
R3
1
=T2
2
R3
2
where T1and R1are the period and distance of the unknown mass star, and T2
and R2are the given period and distance of the planet.
Step 2: We are given that the period T2of the planet is 2 years, so T2= 2
years. The distance R2between the planet and the star is 2×108km or
R2= 2 ×108km.
Step 3: The period T1of the star is unknown and the distance R1is also
unknown. Let the mass of the planet be represented by Mpand the mass of the
star be represented by Ms.
Step 4: We know that the mass of the star Msis directly proportional to
the cube of the distance R1and the square of the period T1. We can rewrite
the formula in terms of mass as:
Ms
Mp
=(R1
R2)3(T1
T2)2
Step 5: Substituting the given values T2= 2 years and R2= 2 ×108km into
the equation above, we get:
Ms
Mp
=(R1
2×108)3(T1
2)2
Step 6: Simplifying, we find that:
Ms
Mp
=(R1
2×108)3(T1
2)2
= 1
Step 7: Therefore, the mass of the star, in terms of the mass of the planet,
is equal to the ratio of their masses which is 1. Hence, the mass of the star is
equal to the mass of the planet.
Question 12
Question
Two planets, Planet A and Planet B, orbit a star. Planet A has a semi-major
axis of 3 AU and an orbital period of 4 years. Planet B has a semi-major axis
of 4 AU and an orbital period of 6 years. Determine which planet has a greater
orbital speed and by how much.
11
Solution
Step 1: Use Kepler’s Third Law to find the orbital speed of each planet. Ke-
pler’s Third Law states that the square of the orbital period (T) of a planet is
proportional to the cube of the semi-major axis (a) of its orbit. Mathematically,
this can be expressed as:
T2
A
a3
A
=T2
B
a3
B
where TAand aAare the orbital period and semi-major axis of Planet A, while
TBand aBare the orbital period and semi-major axis of Planet B.
Step 2: Using the given values for Planet A: TA= 4 years, aA= 3 AU.
Substitute these values into the equation:
42
33=16
27 =T2
B
a3
B
Step 3: Using the given values for Planet B: TB= 6 years, aB= 4 AU. Solve
for T2
B:
T2
B=16
27 ×a3
B=16
27 ×43=16
27 ×64
T2
B=16 ×64
27
T2
B=1024
27
Step 4: Now, find the orbital speed vof each planet using the formula:
v=2πa
T
where vis the orbital speed, ais the semi-major axis, and Tis the orbital period.
Step 5: For Planet A: a= 3 AU, T= 4 years.
vA=2π×3
4=6π
4
vA=3π
2
Step 6: For Planet B: a= 4 AU, T= 6 years.
vB=2π×4
6=8π
6
vB=4π
3
Step 7: Compare the orbital speeds of the two planets: vA=3π
2and vB=
4π
3. Thus, the greater orbital speed is vB=4π
3AU/year.
Therefore, Planet B has a greater orbital speed than Planet A by π
6AU/year.
12
Question 13
Question
According to Kepler’s third law of planetary motion, the square of the orbital
period of a planet is proportional to the cube of its semi-major axis. Given that
the orbital period of Earth is 365.25 days and its average distance from the Sun
is approximately 1 astronomical unit (AU), determine the orbital period of a
hypothetical planet with an average distance from the Sun of 3 AU.
Solution
Step 1: Write down Kepler’s third law in mathematical form:
Kepler’s third law states that the ratio of the squares of the periods of
any two planets is equal to the ratio of the cubes of their semi-major axes.
Mathematically, this can be expressed as:
(T1
T2)2
=(a1
a2)3
where T1and T2are the periods of the two planets, and a1and a2are their
semi-major axes.
Step 2: Given that the orbital period of Earth (T1) is 365.25 days and its
average distance from the Sun (a1) is 1 AU, we are asked to find the orbital
period of a planet with an average distance from the Sun of 3 AU (a2).
Step 3: Substitute the known values into Kepler’s third law equation:
(365.25
T2)2
=(1
3)3
Step 4: Solve for T2:
(365.25
T2)2
=(1
27)
365.25
T2
=√1
27
365.25
T2
=1
3
3×365.25 = T2
T2= 1095.75 days
Therefore, the orbital period of the hypothetical planet with an average
distance from the Sun of 3 AU is approximately 1095.75 days.
13
Question 14
Question
Kepler’s second law states that a planet moves fastest when it is closest to the
Sun (at perihelion) and slowest when it is farthest from the Sun (at aphelion).
Suppose a planet has a perihelion distance of 0.5 AU and an aphelion distance
of 1.5 AU. Calculate the average speed of the planet in its elliptical orbit in
terms of its speed at perihelion.
Solution
Step 1: Let’s denote the planet’s speed at perihelion as vp. The planet covers
equal areas in equal times, so we have:
Area swept at perihelion =Area swept at aphelion
Step 2: The area swept by the planet at perihelion can be calculated using the
formula for the area of an ellipse:
Area =π×semi-major axis ×semi-minor axis
Thus, at perihelion:
π×0.5×b=π×0.5vp
Step 3: Similarly, at aphelion:
π×1.5×b=π×1.5vp
⇒1.5×b= 0.5×b
⇒b=1
3b
Step 4: The average speed of the planet in its orbit can be calculated using the
formula:
Average speed =total distance
total time
Step 5: The total distance covered by the planet in one complete orbit is the
circumference of an ellipse, given by the formula:
2π×√a2+b2
2
where ais the semi-major axis and bis the semi-minor axis. Step 6: The
total time to complete one elliptical orbit is the same whether the planet is at
perihelion or aphelion. The time taken to travel from perihelion to aphelion
is equal to the time taken to travel from aphelion back to perihelion. Step 7:
Using the equations from Steps 2 and 3 along with the information from Step 5,
we can calculate the average speed of the planet in its elliptical orbit in terms
of its speed at perihelion.
14
Question 15
Question
The asteroid Eris has an average orbital radius from the Sun of approximately
10 billion kilometers. If the orbital period of Eris is about 560 years, what is
the mass of the Sun? (Assume Eris follows a circular orbit.)
Solution
Step 1: We can use Kepler’s Third Law of Planetary Motion to relate the orbital
period of a planet to its average orbital radius. Kepler’s Third Law states that
the square of the orbital period of a planet is directly proportional to the cube
of the average orbital radius. Mathematically, this can be written as:
T2=k×R3
where Tis the orbital period, Ris the average orbital radius, and kis a constant
of proportionality. Step 2: We can rearrange the equation to solve for the
constant k:
k=T2
R3
Step 3: Since the asteroid Eris has an average orbital radius of 10 billion kilo-
meters (or 10 ×109km) and an orbital period of 560 years, we can substitute
these values into the equation to find the constant k:
k=(560 years)2
(10 ×109km)3
Step 4: Calculate the value of k:
k=313600 years2
109×109×109km3=313600 years2
1027 km3
k= 3.136 ×10−14 years2/km3
Step 5: Now that we know the constant k, we can use it to find the mass of
the Sun using the average orbital radius of the Earth (1.496 ×108km) and the
orbital period of Earth (1 year):
mSun =4π2
k
R3
T2
mSun =4π2
3.136 ×10−14
(10 ×109)3
(560)2
Step 6: Calculate the mass of the Sun:
mSun =4π2
3.136 ×10−14
1027
313600
15
mSun =4×(π)2×1027
3.136 ×313600
mSun ≈2.03 ×1030 kg
Therefore, the mass of the Sun is approximately 2.03 ×1030 kg.
Question 16
Question
Kepler’s Third Law of Planetary Motion states that the square of the orbital
period of a planet is proportional to the cube of its semi-major axis. Suppose a
planet has an orbital period of 10 years and a semi-major axis of 3 astronomical
units (AU). Calculate the orbital period of another planet with a semi-major
axis of 6 AU.
Solution
To apply Kepler’s Third Law of Planetary Motion, we can use the formula:
T2
1
a3
1
=T2
2
a3
2
where: - T1and a1are the orbital period and semi-major axis of the first planet,
respectively, - T2and a2are the orbital period and semi-major axis of the second
planet, respectively.
Step 1: Substitute the values given for the first planet: T1= 10 years and
a1= 3 AU, into the formula.
102
33=T2
2
63
Step 2: Simplify the equation.
100
27 =T2
2
216
Step 3: Cross multiply to find T2
2.
T2
2=100
27 ×216
T2
2=100 ×216
27
T2
2=21600
27
T2
2= 800
16
Step 4: Take the square root of both sides to find T2.
T2=√800
T2= 28.28 years
Therefore, the orbital period of the other planet with a semi-major axis of
6 AU is approximately 28.28 years.
Question 17
Question
Explain Kepler’s Third Law of Planetary Motion and how it relates to the period
and distance of a planet from the Sun.
Solution
Kepler’s Third Law of Planetary Motion states that the square of the period
of revolution of a planet around the Sun is proportional to the cube of the
semi-major axis of its orbit. Mathematically, this can be expressed as:
T2=k·a3
where: - Tis the period of revolution of the planet, - ais the semi-major axis
of the planet’s orbit, - kis a constant of proportionality.
Step 1: Let’s denote the period of revolution of Planet A as TA, the semi-
major axis of its orbit as aA, and the constant of proportionality for Planet A
as kA.
Step 2: Similarly, denote the period of revolution of Planet B as TB, the
semi-major axis of its orbit as aB, and the constant of proportionality for Planet
B as kB.
Step 3: Using Kepler’s Third Law, we have the following equations for
Planets A and B:
T2
A=kA·a3
A
T2
B=kB·a3
B
Step 4: Now, let’s compare the two planets. Since both planets revolve
around the same Sun, we can set kA=kB=kfor simplicity.
Step 5: By setting the two equations equal to each other, we get:
T2
A=T2
B
k·a3
A=k·a3
B
Step 6: Simplifying the above equation, we find that:
a3
A=a3
B
17
Step 7: Taking the cube root of both sides, we get:
aA=aB
Step 8: Therefore, the semi-major axes of the orbits of both planets are
equal. This means that the distance of a planet from the Sun is directly related
to its period of revolution.
Question 18
Question
Kepler’s third law of planetary motion states that the square of the period of
revolution of a planet is proportional to the cube of its average distance from
the Sun. Suppose a newly discovered planet, Planet X, has an average distance
from the Sun that is twice that of Earth. If Earth’s period of revolution around
the Sun is 1 year, what is Planet X’s period of revolution in terms of Earth
years?
Solution
Step 1: Let the average distance of Earth from the Sun be represented as rE,
and the period of revolution of Earth be denoted as TE. Given that the average
distance of Planet X from the Sun is twice that of Earth’s, we have the average
distance of Planet X, rX= 2rE.
Step 2: According to Kepler’s third law, the square of the period of revolution
of a planet is proportional to the cube of its average distance from the Sun.
Mathematically, this relationship is represented as:
(TX
TE)2
=(rX
rE)3
Step 3: Substituting the values for rXand rE, we have:
(TX
TE)2
=(2rE
rE)3
= 23
Step 4: Simplifying the equation, we get:
(TX
TE)2
= 8
Step 5: Taking the square root of both sides, we find:
TX
TE
=√8 = 2√2
Step 6: Therefore, the period of revolution of Planet X in terms of Earth
years is TX= 2√2×TE= 2√2×1 = 2√2years. Thus, the period of revolution
of Planet X is approximately 2.83 Earth years.
18
Question 19
Question
An asteroid orbits the Sun with a semi-major axis of 3.2 AU. If the time taken
for the asteroid to complete one orbit is 5.6 years, determine the mass of the
Sun in terms of the mass of the Earth. Assume the asteroid has a negligible
mass compared to the Sun.
Solution
Step 1: Find the Sun’s mass in terms of the asteroid’s orbital period.
T2=4π2a3
GM
where: T= 5.6years (orbital period), a= 3.2AU (semi-major axis), G=
universal gravitational constant, and M= mass of the Sun.
Step 2: Substitute the given values into the equation.
(5.6)2=4π2(3.2)3
G·M
Step 3: Solve for M.
M=4π2(3.2)3
G·(5.6)2
Step 4: Approximate the value of Min terms of Earth’s mass.
M=4π2(3.2)3
(6.674 ×10−11)·(5.6)2
M≈3.68 ×105M⊕
Therefore, the mass of the Sun in terms of the mass of the Earth is approx-
imately 3.68 ×105times the mass of the Earth.
Question 20
Question
Consider a hypothetical planet with an orbital period of 200 days. If the semi-
major axis of the planet’s orbit is 2 AU (astronomical units), what is the mass
of the central star in solar masses? (Assume the star is much more massive than
the planet.)
19
Solution
Step 1: Recall Kepler’s third law, which relates the orbital period of a planet
(T), the semi-major axis of its orbit (a), and the mass of the central star (M)
using the equation:
T2=4π2
GM a3
where Gis the gravitational constant.
Step 2: We are given that T= 200 days (1day = 24 ×60 ×60 seconds) and
a= 2 AU. We know that 1AU = 1.496 ×108km. Converting afrom AU to
km:
a= 2 AU ×1.496 ×108km/AU = 2.992 ×108km
Step 3: We now have all the values needed to solve for the mass of the
central star. Substituting T,a, and the known constants into Kepler’s third law
equation:
(200 ×24 ×60 ×60)2=4π2
6.67 ×10−11M(2.992 ×108)3
Step 4: Simplify the equation by squaring 200 ×24 ×60 ×60 and cubing
2.992 ×108. Then solve for M:
3.456 ×109=4π2
6.67 ×10−11M2.160 ×1024
Step 5: Rearrange the equation to solve for M:
M=4π2
6.67 ×10−11 ×3.456 ×109(2.160 ×1024)
Step 6: Calculate the value of Musing a calculator to find the mass of the
central star in solar masses.
Question 21
Question
Explain Kepler’s Third Law of Planetary Motion and calculate the orbital period
of a hypothetical planet orbiting a star with a semimajor axis of 3 AU.
Solution
Kepler’s Third Law of Planetary Motion states that the square of the orbital
period (T) of a planet is directly proportional to the cube of the semimajor axis
(a) of its orbit. Mathematically, it can be expressed as:
T2=k×a3
20
where kis a constant that depends on the system of measurement used.
Step 1: Given that the semimajor axis a= 3 AU, we can find the orbital
period Tin years.
Step 2: Since we are dealing with the square of the orbital period and the
cube of the semimajor axis, we need to set up a proportion to solve for T.
T2=k×a3
T2=k×33
T2= 27k
Step 3: Now, if we have information about the Earth’s orbital period and
semimajor axis, we can use those values to find the constant k. Let’s say TEarth =
1year and aEarth = 1 AU. Plugging these values into the equation:
12= 27k
1 = 27k
k=1
27
Step 4: Substituting the value of kback into the equation, we can solve for
the orbital period Tof the hypothetical planet.
T2=1
27 ×33
T2=1
27 ×27
T2= 1
T= 1 year
Therefore, the orbital period of the hypothetical planet orbiting a star with
a semimajor axis of 3 AU would be 1 year.
Question 22
Question
According to Kepler’s Laws of Planetary Motion, a planet moves fastest when
it is closest to the Sun. Suppose a planet takes 200 days to complete one orbit
around the Sun. If the planet is closest to the Sun at a distance of 0.3 AU and
farthest from the Sun at a distance of 0.7 AU, determine the average speed and
the maximum speed of the planet in its orbit (in AU/day).
21
Solution
Step 1: To find the average speed of the planet, we will use the formula for
average speed:
Average Speed =Total Distance
Total Time
Step 2: The total distance the planet travels during one orbit can be cal-
culated using the sum of the distances when the planet is closest and farthest
from the Sun:
Total Distance =Distance when closest +Distance when farthest
Step 3: Given that the planet is closest to the Sun at 0.3 AU and farthest
at 0.7 AU, we have:
Total Distance = 0.3AU + 0.7AU = 1.0AU
Step 4: Given that the planet takes 200 days to complete one orbit, the total
time is 200 days:
Total Time = 200 days
Step 5: Now we can calculate the average speed of the planet:
Average Speed =1.0AU
200 days = 0.005 AU/day
Step 6: To find the maximum speed of the planet, we note that the planet
moves fastest when closest to the Sun. Thus, the maximum speed of the planet
is the speed when it is at 0.3 AU.
Step 7: Since the planet takes 200 days to complete one orbit, at 0.3 AU,
the planet travels a distance of 0.3 AU in half of an orbit (i.e., 100 days).
Max Speed =0.3AU
100 days = 0.003 AU/day
Therefore, the average speed of the planet is 0.005 AU/day, and the maxi-
mum speed of the planet in its orbit is 0.003 AU/day.
Question 23
Question
A planet orbits a star in an elliptical orbit where the distance between the
planet and the star at closest approach is 28 million kilometers and the distance
at farthest approach is 100 million kilometers. If the planet takes 200 days to
complete one orbit around the star, what is the velocity of the planet when it
is at its closest approach to the star?
22
Solution
Step 1: Find the semi-major axis of the planet’s orbit. This can be calculated as
the average of the closest and farthest distances from the star. Let rclosest = 28
million km and rfarthest = 100 million km. The semi-major axis is given by:
a=rclosest +rfarthest
2
a=28 + 100
2= 64 million km = 64 ×106km
Step 2: Calculate the eccentricity of the orbit. The eccentricity of the ellipse
can be calculated as:
e=rfarthest −rclosest
rfarthest +rclosest
Substitute the values:
e=100 −28
100 + 28 =72
128 = 0.5625
Step 3: Use Kepler’s Second Law to find the velocity of the planet at its
closest approach. The law states that the line joining the planet and the sun
sweeps out equal areas in equal intervals of time. At closest approach, the planet
moves fastest. The velocity of the planet at any position can be found using the
equation:
v=√GM(1 + e)
a(1 −e)
where vis the velocity, Gis the gravitational constant, Mis the mass of the
star, ais the semi-major axis, and eis the eccentricity of the orbit. At the
closest approach, r=rclosest:
vclosest =√GM(1 + e)
a(1 −e)
Substitute the given values and constants into the equation:
vclosest =√6.674 ×10−11 ×M(1 + 0.5625)
64 ×106×(1 −0.5625)
vclosest =√6.674 ×10−11 ×M×1.5625
64 ×106×0.4375
vclosest =√10.41875 ×10−11 ×M
28 ×106
vclosest =√0.37245 ×10−5×M
Therefore, the velocity of the planet when it is at its closest approach to the
star is √0.37245 ×10−5×M.
23
Question 24
Question
The orbit of a certain planet is known to be an ellipse with semi-major axis of
2.5×106km. If the planet has an eccentricity of 0.4, determine the distance
between the planet and the Sun at its closest approach.
Solution
Step 1: Recall Kepler’s Second Law, which states that a planet will sweep out
equal areas in equal times as it travels along its orbit. When a planet is at its
closest approach to the Sun (called perihelion), it moves fastest. This means
the line connecting the Sun and the planet sweeps out an equal area in a given
time compared to other points in the orbit.
Step 2: The distance between the planet and the Sun at closest approach
is equal to the semi-minor axis of the elliptical orbit. The semi-major axis and
the eccentricity of the orbit can be used to calculate the semi-minor axis using
the formula:
b=a√1−e2
where ais the semi-major axis, eis the eccentricity, and bis the semi-minor
axis.
Step 3: Plug in the given values:
b= 2.5×106km ×√1−0.42
b= 2.5×106km ×√1−0.16
b= 2.5×106km ×√0.84
b= 2.5×106km ×0.916515
b≈2.2912875 ×106km
Step 4: Therefore, the distance between the planet and the Sun at its closest
approach is approximately 2.291 ×106km.
Question 25
Question
Suppose a planet has an elliptical orbit around the Sun such that its eccentricity
is 0.4. If the distance between the planet and the Sun at its closest approach
(perihelion) is 0.3 astronomical units (AU), determine the distance between the
planet and the Sun at its farthest point (aphelion).
24
Solution
Step 1: Recall that the eccentricity of an elliptical orbit is defined as the ratio
of the distance between the foci of the ellipse to the length of the major axis.
Mathematically, the eccentricity (e) can be expressed as:
e=c
a
where cis half the distance between the foci of the ellipse, and ais half the
length of the major axis.
Step 2: We are given that the eccentricity of the planet’s orbit is 0.4, and
the distance between the planet and the Sun at perihelion is 0.3 AU. This
information allows us to set up an equation to solve for the distance between
the planet and the Sun at aphelion.
Step 3: Let the distance between the planet and the Sun at aphelion be x
AU. From the definition of eccentricity, we have:
0.4 = c
a=x−0.3
x+0.3
2
Step 4: Simplifying the equation, we get:
0.4 = 2(x−0.3)
x+ 0.3
Step 5: Cross multiplying gives:
0.4(x+ 0.3) = 2(x−0.3)
Step 6: Expanding both sides of the equation yields:
0.4x+ 0.12 = 2x−0.6
Step 7: Rearranging the terms gives:
0.12 + 0.6 = 2x−0.4x
0.72 = 1.6x
Step 8: Finally, solving for x, we find:
x=0.72
1.6≈0.45 AU
Step 9: Therefore, the distance between the planet and the Sun at its farthest
point (aphelion) is approximately 0.45 AU.
25
Step 5: To find T2
2, we can cross multiply:
144 ×3375 = 125 ×T2
2
Step 6: Solving for T2
2gives:
T2
2=144 ×3375
125
Step 7: Calculate the value of T2:
T2=√144 ×3375
125
Step 8: Simplify the expression to find the approximate orbital period of the
second planet.
Question 2
Question
Kepler’s third law of planetary motion states that the square of the period of
revolution (T, in years) of a planet is directly proportional to the cube of the
semi-major axis of its orbit (a, in astronomical units). The proportionality can
be expressed as T2=k·a3, where kis a constant. Consider two planets, A and
B, with periods of revolution of 5 years and 10 years, respectively. If planet A
has a semi-major axis of orbit of 2 astronomical units, what is the semi-major
axis of orbit for planet B?
Solution
Step 1: Calculate the value of kfor each planet using Kepler’s third law. Step
2: Use the value of kand the given period of revolution for planet B to find the
semi-major axis of its orbit.
Step 1: For planet A:
T2
A=k·a3
A
52=k·23
25 = 8k
k=25
8= 3.125
For planet B:
T2
B=k·a3
B
102= 3.125 ·a3
B
100 = 3.125a3
B
a3
B=100
3.125 = 32
aB=3
√32 = 2
Therefore, the semi-major axis of orbit for planet B is 2 astronomical units.
2
Question 3
Question
Consider a planet with a semi-major axis of 2 AU orbiting a star with a mass
of 2×1030 kg. If the planet completes one full orbit in 2 years, determine its
orbital period, orbital speed, and gravitational force between the planet and the
star.
Solution
Step 1: Calculate the orbital period Tusing Kepler’s Third Law of Planetary
Motion:
T2=4π2
GM a3,
where T= orbital period, G= gravitational constant (6.67 ×10−11 m3/kg ·s2),
M= mass of the star (2×1030 kg), a= semi-major axis of the planet’s orbit
(2 AU).
Step 2: Substitute the given values into the equation to solve for T:
T2=4π2
(6.67 ×10−11)(2 ×1030)(2 AU)3
T2=4π2
1.334 ×1020 (8) ≈5.98 ×108
T≈√5.98 ×108≈24,464 days
Therefore, the orbital period of the planet is approximately 24,464 days.
Step 3: Calculate the orbital speed vof the planet using the formula:
v=2πa
T,
where v= orbital speed, a= semi-major axis of the planet’s orbit (2 AU), T=
orbital period (24,464 days).
Step 4: Substitute the values into the equation to find v:
v=2π(2)
24,464 ≈4π
24,464 ≈0.000257 AU/day
Therefore, the orbital speed of the planet is approximately 0.000257 AU/day.
Step 5: Calculate the gravitational force Fbetween the planet and the star
using Newton’s Law of Universal Gravitation:
F=GMm
r2,
where F= gravitational force, G= gravitational constant (6.67 ×10−11 m3/kg ·
s2), M= mass of the star (2×1030 kg), m= mass of the planet, r= distance
between the planet and the star.
3
Step 6: Assume the mass of the planet is negligible compared to the star,
then m≈0. The distance rcan be considered approximately equal to the
semi-major axis a.
Step 7: Substitute the values into the equation to find F:
F=(6.67 ×10−11)(2 ×1030)(0)
(2 AU)2= 0
Therefore, the gravitational force between the planet and the star is approx-
imately 0 N.
Question 4
Question
Given that a planet has an orbit with a semi-major axis of 2.5 AU, determine
the period of its orbit in years. Assume the planet’s orbit is nearly circular.
Solution
Step 1: Recall Kepler’s Third Law of Planetary Motion, which states that the
square of the period of a planet’s orbit (T) is proportional to the cube of the
semi-major axis of its orbit (a):
T2∝a3
Step 2: We can write the equation for Kepler’s Third Law as:
T2=k·a3
where kis the constant of proportionality.
Step 3: By rearranging the equation, we find:
T=√k·a3
1
Step 4: To determine the period of the planet’s orbit in years, we need to
find the value of the constant of proportionality (k). Since the planet’s orbit is
nearly circular, we can use Earth as a reference point. Earth has a semi-major
axis of 1 AU and a period of 1 year.
Step 5: Substituting T= 1 year and a= 1 AU into the equation gives:
1 = k·13
1 = k
Step 6: Now, we can plug k= 1 into the equation for the planet’s orbit:
T=√1·2.53
T=√15.625
T= 3.953 years
Therefore, the period of the planet’s orbit is approximately 3.953 years.
4
Question 5
Question
An exoplanet orbits a star with a semi-major axis of 0.25 astronomical units. If
the period of the exoplanet is 150 days, determine the mass of the star. (Hint:
Use Kepler’s third law).
Solution
Step 1: Recall Kepler’s third law which states that the square of the period (T)
of a planet is directly proportional to the cube of the semi-major axis (a) of its
orbit. Mathematically, this can be expressed as:
T2=k·a3
where kis a constant.
Step 2: We can rewrite the equation in terms of kby considering the data
given for the exoplanet:
T2
planet =k·a3
planet
(150 days)2=k·(0.25 AU)3
Step 3: Solve for the constant k:
(1502) = k·(0.253)
k=1502
0.253
Step 4: Now, using the mass of the star (Mstar) and the period of the star
(Tstar) in the modified Kepler’s third law, we have:
T2
star =k·a3
star
T2
star =1502
0.253·a3
star
Step 5: Given that astar = 0 for a star, the equation simplifies to:
T2
star =1502
0.253·0
Tstar = 0
Step 6: Since the period of a star is negligible, the mass of the star does not
affect the motion of the exoplanet.
Step 7: Therefore, the mass of the star is not required to determine the
exoplanet’s motion.
5
Question 6
Question
Suppose a spacecraft is in a circular orbit around a planet with a period of
20 hours. If the spacecraft’s altitude above the planet’s surface is 4,000 km,
determine the mass of the planet assuming the radius of the planet is 6,000 km.
Solution
Let’s assume that the planet can be approximated as a sphere. In a circular
orbit, the gravitational force provides the necessary centripetal force to keep the
spacecraft in its orbit. We can use Kepler’s third law: T2=4π2r3
GM , where: - T
is the period of the orbit, - ris the distance from the center of the planet to the
center of the spacecraft (altitude + radius of the planet), - Gis the gravitational
constant, - Mis the mass of the planet.
Step 1: Determine distance rfrom the center of the planet to the
center of the spacecraft
• Altitude above the planet’s surface: 4,000 km
• Radius of the planet: 6,000 km
Therefore, r=altitude +radius = 4,000 km + 6,000 km = 10,000 km.
Step 2: Calculate the period Tin seconds
• Given period T= 20 hours
Convert 20 hours to seconds: 20 ×60 ×60 = 72,000 s.
Step 3: Plug the known values into Kepler’s third law and solve
for the planet’s mass M
T2=4π2r3
GM
(72,000 s)2=4π2(10,000 km)3
G·M
5.184 ×109s2=4π2×1×109km3
G×M
Step 4: Solve for the mass Mof the planet
M=4π2×1×109km3
5.184 ×109s2×G
M=4π2×1×109km3
5.184 ×109s2×6.67430 ×10−11 m3kg−1s−2
M≈3.4×1022 kg
Therefore, the mass of the planet is approximately 3.4×1022 kg.
6
Question 7
Question
At a certain point in its orbit, a planet is 0.8 AU from the Sun and moving at a
speed of 28 km/s. Determine the shortest distance of the planet from the Sun
and the speed of the planet at this point in its orbit.
Given: a= 0.8AU, v= 28 km/s
Solution
Step 1: Determine the eccentricity of the planet’s orbit using Kepler’s Third
Law. T2
a3=4π2
GM
T2
(0.8)3=4π2
GM
T2
0.512 =4π2
GM
Solving for Tyields:
T=√0.512 ·4π2
GM
Step 2: Use the information provided to determine the instantaneous speed
and distance at the shortest distance from the Sun.
v2=GM (2
r−1
a)
Plug in the given values and solve for r:
282=GM (2
r−1
0.8)
Step 3: Determine the speed of the planet at its closest distance from the
Sun.
v′=√GM (2
r−1
a)
Substitute the value of robtained from step 2 and solve for v′.
Question 8
Question
According to Kepler’s third law of planetary motion, the square of the orbital
period of a planet is directly proportional to the cube of the semi-major axis of
7
its orbit. Suppose Planet X has an orbital period of 5 years and a semi-major
axis of 3 AU (astronomical units). Calculate the orbital period of Planet Y if it
has a semi-major axis of 8 AU.
Solution
Step 1: Write down Kepler’s third law in equation form:
T2∝a3
where Tis the orbital period and ais the semi-major axis.
Step 2: Use the given values for Planet X (T= 5 years, a= 3 AU) to set up
a proportion:
T2
1
a3
1
=T2
2
a3
2
Step 3: Substitute the values for Planet X into the proportion:
52
33=T2
2
83
Step 4: Simplify the equation:
25
27 =T2
2
512
Step 5: Cross multiply to solve for T2:
25 ×512 = 27 ×T2
2
T2
2=25 ×512
27
Step 6: Calculate T2:
T2
2=12800
27
T2=√12800
27
T2≈15.07 years
Therefore, the orbital period of Planet Y is approximately 15.07 years.
Question 9
Question
Given that the semi-major axis of an exoplanet’s orbit is 2.5 AU, determine the
period of its orbit in years.
8
Solution
Step 1: Recall Kepler’s Third Law of Planetary Motion:
T2=k×a3
where: T= period of orbit (in years) a= semi-major axis of the orbit (in AU)
k= proportionality constant
Step 2: Substitute the given semi-major axis (a= 2.5AU) into the equation.
T2=k×(2.5)3
Step 3: Since kis a constant, we can rewrite the equation as:
T2=k×15.625
Step 4: To solve for T, we need to find the value of the constant k. This
can be done by using data from a known system, such as the Earth’s orbit with
a= 1 AU and T= 1 year.
1 = k×13
k= 1
Step 5: Substitute k= 1 back into the equation:
T2= 15.625
Step 6: Take the square root of both sides to solve for T:
T=√15.625 = 3.944
Therefore, the period of the exoplanet’s orbit is approximately 3.944 years.
Question 10
Question
An exoplanet orbits its star in a period of 30 days at an average distance of 0.1
AU. If the mass of the star is known to be 2×1030 kg, determine the mass of
the exoplanet.
Solution
Step 1: We can use Kepler’s third law of planetary motion to determine the
mass of the exoplanet. Kepler’s third law states that the square of the orbital
period of a planet is directly proportional to the cube of the semi-major axis of
its orbit. Mathematically, this can be expressed as:
T2
1
T2
2
=a3
1
a3
2
9
where T1and T2are the orbital periods of the two planets, and a1and a2are
their semi-major axes.
Step 2: Given that the exoplanet’s orbital period is 30 days and its average
distance is 0.1 AU, and that the star’s mass is 2×1030 kg, we can set up the
equation:
302
T2
2
=0.13
a3
2
Step 3: To solve for the mass of the exoplanet, we need to find its semi-
major axis. Since the exoplanet orbits the star, the semi-major axis is the
average distance between the exoplanet and the star:
semi-major axis =average distance
Step 4: Substituting the given values into the equation, we get:
302
T2
2
=0.13
0.13
Step 5: Simplifying the equation further, we find:
900
T2
2
= 1
Step 6: Solving for the orbital period of the exoplanet, we get:
T2
2= 900 ⇒T2= 30 days
Step 7: Now that we have the orbital period of the exoplanet, we can use
Kepler’s third law to find its mass. Kepler’s third law in this case simplifies to:
M1=(T1
T2)2
×M2
Step 8: Substituting the known values into the equation, we can now solve
for the mass of the exoplanet:
M1=(30
30)2
×2×1030 = 2 ×1030 kg
Step 9: Therefore, the mass of the exoplanet is 2×1030 kg.
Question 11
Question
Consider a planet in a circular orbit around a star. The planet takes 2 years to
complete one orbit. If the distance between the planet and the star is 2×108
km, calculate the mass of the star in terms of the mass of the planet.
10
Solution
Step 1: We can start by using Kepler’s third law of planetary motion which
states that the square of the period of revolution of a planet is directly propor-
tional to the cube of its mean distance from the Sun. Mathematically, this can
be expressed as:
T2
1
R3
1
=T2
2
R3
2
where T1and R1are the period and distance of the unknown mass star, and T2
and R2are the given period and distance of the planet.
Step 2: We are given that the period T2of the planet is 2 years, so T2= 2
years. The distance R2between the planet and the star is 2×108km or
R2= 2 ×108km.
Step 3: The period T1of the star is unknown and the distance R1is also
unknown. Let the mass of the planet be represented by Mpand the mass of the
star be represented by Ms.
Step 4: We know that the mass of the star Msis directly proportional to
the cube of the distance R1and the square of the period T1. We can rewrite
the formula in terms of mass as:
Ms
Mp
=(R1
R2)3(T1
T2)2
Step 5: Substituting the given values T2= 2 years and R2= 2 ×108km into
the equation above, we get:
Ms
Mp
=(R1
2×108)3(T1
2)2
Step 6: Simplifying, we find that:
Ms
Mp
=(R1
2×108)3(T1
2)2
= 1
Step 7: Therefore, the mass of the star, in terms of the mass of the planet,
is equal to the ratio of their masses which is 1. Hence, the mass of the star is
equal to the mass of the planet.
Question 12
Question
Two planets, Planet A and Planet B, orbit a star. Planet A has a semi-major
axis of 3 AU and an orbital period of 4 years. Planet B has a semi-major axis
of 4 AU and an orbital period of 6 years. Determine which planet has a greater
orbital speed and by how much.
11
Solution
Step 1: Use Kepler’s Third Law to find the orbital speed of each planet. Ke-
pler’s Third Law states that the square of the orbital period (T) of a planet is
proportional to the cube of the semi-major axis (a) of its orbit. Mathematically,
this can be expressed as:
T2
A
a3
A
=T2
B
a3
B
where TAand aAare the orbital period and semi-major axis of Planet A, while
TBand aBare the orbital period and semi-major axis of Planet B.
Step 2: Using the given values for Planet A: TA= 4 years, aA= 3 AU.
Substitute these values into the equation:
42
33=16
27 =T2
B
a3
B
Step 3: Using the given values for Planet B: TB= 6 years, aB= 4 AU. Solve
for T2
B:
T2
B=16
27 ×a3
B=16
27 ×43=16
27 ×64
T2
B=16 ×64
27
T2
B=1024
27
Step 4: Now, find the orbital speed vof each planet using the formula:
v=2πa
T
where vis the orbital speed, ais the semi-major axis, and Tis the orbital period.
Step 5: For Planet A: a= 3 AU, T= 4 years.
vA=2π×3
4=6π
4
vA=3π
2
Step 6: For Planet B: a= 4 AU, T= 6 years.
vB=2π×4
6=8π
6
vB=4π
3
Step 7: Compare the orbital speeds of the two planets: vA=3π
2and vB=
4π
3. Thus, the greater orbital speed is vB=4π
3AU/year.
Therefore, Planet B has a greater orbital speed than Planet A by π
6AU/year.
12
Question 13
Question
According to Kepler’s third law of planetary motion, the square of the orbital
period of a planet is proportional to the cube of its semi-major axis. Given that
the orbital period of Earth is 365.25 days and its average distance from the Sun
is approximately 1 astronomical unit (AU), determine the orbital period of a
hypothetical planet with an average distance from the Sun of 3 AU.
Solution
Step 1: Write down Kepler’s third law in mathematical form:
Kepler’s third law states that the ratio of the squares of the periods of
any two planets is equal to the ratio of the cubes of their semi-major axes.
Mathematically, this can be expressed as:
(T1
T2)2
=(a1
a2)3
where T1and T2are the periods of the two planets, and a1and a2are their
semi-major axes.
Step 2: Given that the orbital period of Earth (T1) is 365.25 days and its
average distance from the Sun (a1) is 1 AU, we are asked to find the orbital
period of a planet with an average distance from the Sun of 3 AU (a2).
Step 3: Substitute the known values into Kepler’s third law equation:
(365.25
T2)2
=(1
3)3
Step 4: Solve for T2:
(365.25
T2)2
=(1
27)
365.25
T2
=√1
27
365.25
T2
=1
3
3×365.25 = T2
T2= 1095.75 days
Therefore, the orbital period of the hypothetical planet with an average
distance from the Sun of 3 AU is approximately 1095.75 days.
13
Question 14
Question
Kepler’s second law states that a planet moves fastest when it is closest to the
Sun (at perihelion) and slowest when it is farthest from the Sun (at aphelion).
Suppose a planet has a perihelion distance of 0.5 AU and an aphelion distance
of 1.5 AU. Calculate the average speed of the planet in its elliptical orbit in
terms of its speed at perihelion.
Solution
Step 1: Let’s denote the planet’s speed at perihelion as vp. The planet covers
equal areas in equal times, so we have:
Area swept at perihelion =Area swept at aphelion
Step 2: The area swept by the planet at perihelion can be calculated using the
formula for the area of an ellipse:
Area =π×semi-major axis ×semi-minor axis
Thus, at perihelion:
π×0.5×b=π×0.5vp
Step 3: Similarly, at aphelion:
π×1.5×b=π×1.5vp
⇒1.5×b= 0.5×b
⇒b=1
3b
Step 4: The average speed of the planet in its orbit can be calculated using the
formula:
Average speed =total distance
total time
Step 5: The total distance covered by the planet in one complete orbit is the
circumference of an ellipse, given by the formula:
2π×√a2+b2
2
where ais the semi-major axis and bis the semi-minor axis. Step 6: The
total time to complete one elliptical orbit is the same whether the planet is at
perihelion or aphelion. The time taken to travel from perihelion to aphelion
is equal to the time taken to travel from aphelion back to perihelion. Step 7:
Using the equations from Steps 2 and 3 along with the information from Step 5,
we can calculate the average speed of the planet in its elliptical orbit in terms
of its speed at perihelion.
14
Question 15
Question
The asteroid Eris has an average orbital radius from the Sun of approximately
10 billion kilometers. If the orbital period of Eris is about 560 years, what is
the mass of the Sun? (Assume Eris follows a circular orbit.)
Solution
Step 1: We can use Kepler’s Third Law of Planetary Motion to relate the orbital
period of a planet to its average orbital radius. Kepler’s Third Law states that
the square of the orbital period of a planet is directly proportional to the cube
of the average orbital radius. Mathematically, this can be written as:
T2=k×R3
where Tis the orbital period, Ris the average orbital radius, and kis a constant
of proportionality. Step 2: We can rearrange the equation to solve for the
constant k:
k=T2
R3
Step 3: Since the asteroid Eris has an average orbital radius of 10 billion kilo-
meters (or 10 ×109km) and an orbital period of 560 years, we can substitute
these values into the equation to find the constant k:
k=(560 years)2
(10 ×109km)3
Step 4: Calculate the value of k:
k=313600 years2
109×109×109km3=313600 years2
1027 km3
k= 3.136 ×10−14 years2/km3
Step 5: Now that we know the constant k, we can use it to find the mass of
the Sun using the average orbital radius of the Earth (1.496 ×108km) and the
orbital period of Earth (1 year):
mSun =4π2
k
R3
T2
mSun =4π2
3.136 ×10−14
(10 ×109)3
(560)2
Step 6: Calculate the mass of the Sun:
mSun =4π2
3.136 ×10−14
1027
313600
15
mSun =4×(π)2×1027
3.136 ×313600
mSun ≈2.03 ×1030 kg
Therefore, the mass of the Sun is approximately 2.03 ×1030 kg.
Question 16
Question
Kepler’s Third Law of Planetary Motion states that the square of the orbital
period of a planet is proportional to the cube of its semi-major axis. Suppose a
planet has an orbital period of 10 years and a semi-major axis of 3 astronomical
units (AU). Calculate the orbital period of another planet with a semi-major
axis of 6 AU.
Solution
To apply Kepler’s Third Law of Planetary Motion, we can use the formula:
T2
1
a3
1
=T2
2
a3
2
where: - T1and a1are the orbital period and semi-major axis of the first planet,
respectively, - T2and a2are the orbital period and semi-major axis of the second
planet, respectively.
Step 1: Substitute the values given for the first planet: T1= 10 years and
a1= 3 AU, into the formula.
102
33=T2
2
63
Step 2: Simplify the equation.
100
27 =T2
2
216
Step 3: Cross multiply to find T2
2.
T2
2=100
27 ×216
T2
2=100 ×216
27
T2
2=21600
27
T2
2= 800
16
Step 4: Take the square root of both sides to find T2.
T2=√800
T2= 28.28 years
Therefore, the orbital period of the other planet with a semi-major axis of
6 AU is approximately 28.28 years.
Question 17
Question
Explain Kepler’s Third Law of Planetary Motion and how it relates to the period
and distance of a planet from the Sun.
Solution
Kepler’s Third Law of Planetary Motion states that the square of the period
of revolution of a planet around the Sun is proportional to the cube of the
semi-major axis of its orbit. Mathematically, this can be expressed as:
T2=k·a3
where: - Tis the period of revolution of the planet, - ais the semi-major axis
of the planet’s orbit, - kis a constant of proportionality.
Step 1: Let’s denote the period of revolution of Planet A as TA, the semi-
major axis of its orbit as aA, and the constant of proportionality for Planet A
as kA.
Step 2: Similarly, denote the period of revolution of Planet B as TB, the
semi-major axis of its orbit as aB, and the constant of proportionality for Planet
B as kB.
Step 3: Using Kepler’s Third Law, we have the following equations for
Planets A and B:
T2
A=kA·a3
A
T2
B=kB·a3
B
Step 4: Now, let’s compare the two planets. Since both planets revolve
around the same Sun, we can set kA=kB=kfor simplicity.
Step 5: By setting the two equations equal to each other, we get:
T2
A=T2
B
k·a3
A=k·a3
B
Step 6: Simplifying the above equation, we find that:
a3
A=a3
B
17
Step 7: Taking the cube root of both sides, we get:
aA=aB
Step 8: Therefore, the semi-major axes of the orbits of both planets are
equal. This means that the distance of a planet from the Sun is directly related
to its period of revolution.
Question 18
Question
Kepler’s third law of planetary motion states that the square of the period of
revolution of a planet is proportional to the cube of its average distance from
the Sun. Suppose a newly discovered planet, Planet X, has an average distance
from the Sun that is twice that of Earth. If Earth’s period of revolution around
the Sun is 1 year, what is Planet X’s period of revolution in terms of Earth
years?
Solution
Step 1: Let the average distance of Earth from the Sun be represented as rE,
and the period of revolution of Earth be denoted as TE. Given that the average
distance of Planet X from the Sun is twice that of Earth’s, we have the average
distance of Planet X, rX= 2rE.
Step 2: According to Kepler’s third law, the square of the period of revolution
of a planet is proportional to the cube of its average distance from the Sun.
Mathematically, this relationship is represented as:
(TX
TE)2
=(rX
rE)3
Step 3: Substituting the values for rXand rE, we have:
(TX
TE)2
=(2rE
rE)3
= 23
Step 4: Simplifying the equation, we get:
(TX
TE)2
= 8
Step 5: Taking the square root of both sides, we find:
TX
TE
=√8 = 2√2
Step 6: Therefore, the period of revolution of Planet X in terms of Earth
years is TX= 2√2×TE= 2√2×1 = 2√2years. Thus, the period of revolution
of Planet X is approximately 2.83 Earth years.
18
Question 19
Question
An asteroid orbits the Sun with a semi-major axis of 3.2 AU. If the time taken
for the asteroid to complete one orbit is 5.6 years, determine the mass of the
Sun in terms of the mass of the Earth. Assume the asteroid has a negligible
mass compared to the Sun.
Solution
Step 1: Find the Sun’s mass in terms of the asteroid’s orbital period.
T2=4π2a3
GM
where: T= 5.6years (orbital period), a= 3.2AU (semi-major axis), G=
universal gravitational constant, and M= mass of the Sun.
Step 2: Substitute the given values into the equation.
(5.6)2=4π2(3.2)3
G·M
Step 3: Solve for M.
M=4π2(3.2)3
G·(5.6)2
Step 4: Approximate the value of Min terms of Earth’s mass.
M=4π2(3.2)3
(6.674 ×10−11)·(5.6)2
M≈3.68 ×105M⊕
Therefore, the mass of the Sun in terms of the mass of the Earth is approx-
imately 3.68 ×105times the mass of the Earth.
Question 20
Question
Consider a hypothetical planet with an orbital period of 200 days. If the semi-
major axis of the planet’s orbit is 2 AU (astronomical units), what is the mass
of the central star in solar masses? (Assume the star is much more massive than
the planet.)
19
Solution
Step 1: Recall Kepler’s third law, which relates the orbital period of a planet
(T), the semi-major axis of its orbit (a), and the mass of the central star (M)
using the equation:
T2=4π2
GM a3
where Gis the gravitational constant.
Step 2: We are given that T= 200 days (1day = 24 ×60 ×60 seconds) and
a= 2 AU. We know that 1AU = 1.496 ×108km. Converting afrom AU to
km:
a= 2 AU ×1.496 ×108km/AU = 2.992 ×108km
Step 3: We now have all the values needed to solve for the mass of the
central star. Substituting T,a, and the known constants into Kepler’s third law
equation:
(200 ×24 ×60 ×60)2=4π2
6.67 ×10−11M(2.992 ×108)3
Step 4: Simplify the equation by squaring 200 ×24 ×60 ×60 and cubing
2.992 ×108. Then solve for M:
3.456 ×109=4π2
6.67 ×10−11M2.160 ×1024
Step 5: Rearrange the equation to solve for M:
M=4π2
6.67 ×10−11 ×3.456 ×109(2.160 ×1024)
Step 6: Calculate the value of Musing a calculator to find the mass of the
central star in solar masses.
Question 21
Question
Explain Kepler’s Third Law of Planetary Motion and calculate the orbital period
of a hypothetical planet orbiting a star with a semimajor axis of 3 AU.
Solution
Kepler’s Third Law of Planetary Motion states that the square of the orbital
period (T) of a planet is directly proportional to the cube of the semimajor axis
(a) of its orbit. Mathematically, it can be expressed as:
T2=k×a3
20
where kis a constant that depends on the system of measurement used.
Step 1: Given that the semimajor axis a= 3 AU, we can find the orbital
period Tin years.
Step 2: Since we are dealing with the square of the orbital period and the
cube of the semimajor axis, we need to set up a proportion to solve for T.
T2=k×a3
T2=k×33
T2= 27k
Step 3: Now, if we have information about the Earth’s orbital period and
semimajor axis, we can use those values to find the constant k. Let’s say TEarth =
1year and aEarth = 1 AU. Plugging these values into the equation:
12= 27k
1 = 27k
k=1
27
Step 4: Substituting the value of kback into the equation, we can solve for
the orbital period Tof the hypothetical planet.
T2=1
27 ×33
T2=1
27 ×27
T2= 1
T= 1 year
Therefore, the orbital period of the hypothetical planet orbiting a star with
a semimajor axis of 3 AU would be 1 year.
Question 22
Question
According to Kepler’s Laws of Planetary Motion, a planet moves fastest when
it is closest to the Sun. Suppose a planet takes 200 days to complete one orbit
around the Sun. If the planet is closest to the Sun at a distance of 0.3 AU and
farthest from the Sun at a distance of 0.7 AU, determine the average speed and
the maximum speed of the planet in its orbit (in AU/day).
21
Solution
Step 1: To find the average speed of the planet, we will use the formula for
average speed:
Average Speed =Total Distance
Total Time
Step 2: The total distance the planet travels during one orbit can be cal-
culated using the sum of the distances when the planet is closest and farthest
from the Sun:
Total Distance =Distance when closest +Distance when farthest
Step 3: Given that the planet is closest to the Sun at 0.3 AU and farthest
at 0.7 AU, we have:
Total Distance = 0.3AU + 0.7AU = 1.0AU
Step 4: Given that the planet takes 200 days to complete one orbit, the total
time is 200 days:
Total Time = 200 days
Step 5: Now we can calculate the average speed of the planet:
Average Speed =1.0AU
200 days = 0.005 AU/day
Step 6: To find the maximum speed of the planet, we note that the planet
moves fastest when closest to the Sun. Thus, the maximum speed of the planet
is the speed when it is at 0.3 AU.
Step 7: Since the planet takes 200 days to complete one orbit, at 0.3 AU,
the planet travels a distance of 0.3 AU in half of an orbit (i.e., 100 days).
Max Speed =0.3AU
100 days = 0.003 AU/day
Therefore, the average speed of the planet is 0.005 AU/day, and the maxi-
mum speed of the planet in its orbit is 0.003 AU/day.
Question 23
Question
A planet orbits a star in an elliptical orbit where the distance between the
planet and the star at closest approach is 28 million kilometers and the distance
at farthest approach is 100 million kilometers. If the planet takes 200 days to
complete one orbit around the star, what is the velocity of the planet when it
is at its closest approach to the star?
22
Solution
Step 1: Find the semi-major axis of the planet’s orbit. This can be calculated as
the average of the closest and farthest distances from the star. Let rclosest = 28
million km and rfarthest = 100 million km. The semi-major axis is given by:
a=rclosest +rfarthest
2
a=28 + 100
2= 64 million km = 64 ×106km
Step 2: Calculate the eccentricity of the orbit. The eccentricity of the ellipse
can be calculated as:
e=rfarthest −rclosest
rfarthest +rclosest
Substitute the values:
e=100 −28
100 + 28 =72
128 = 0.5625
Step 3: Use Kepler’s Second Law to find the velocity of the planet at its
closest approach. The law states that the line joining the planet and the sun
sweeps out equal areas in equal intervals of time. At closest approach, the planet
moves fastest. The velocity of the planet at any position can be found using the
equation:
v=√GM(1 + e)
a(1 −e)
where vis the velocity, Gis the gravitational constant, Mis the mass of the
star, ais the semi-major axis, and eis the eccentricity of the orbit. At the
closest approach, r=rclosest:
vclosest =√GM(1 + e)
a(1 −e)
Substitute the given values and constants into the equation:
vclosest =√6.674 ×10−11 ×M(1 + 0.5625)
64 ×106×(1 −0.5625)
vclosest =√6.674 ×10−11 ×M×1.5625
64 ×106×0.4375
vclosest =√10.41875 ×10−11 ×M
28 ×106
vclosest =√0.37245 ×10−5×M
Therefore, the velocity of the planet when it is at its closest approach to the
star is √0.37245 ×10−5×M.
23
Question 24
Question
The orbit of a certain planet is known to be an ellipse with semi-major axis of
2.5×106km. If the planet has an eccentricity of 0.4, determine the distance
between the planet and the Sun at its closest approach.
Solution
Step 1: Recall Kepler’s Second Law, which states that a planet will sweep out
equal areas in equal times as it travels along its orbit. When a planet is at its
closest approach to the Sun (called perihelion), it moves fastest. This means
the line connecting the Sun and the planet sweeps out an equal area in a given
time compared to other points in the orbit.
Step 2: The distance between the planet and the Sun at closest approach
is equal to the semi-minor axis of the elliptical orbit. The semi-major axis and
the eccentricity of the orbit can be used to calculate the semi-minor axis using
the formula:
b=a√1−e2
where ais the semi-major axis, eis the eccentricity, and bis the semi-minor
axis.
Step 3: Plug in the given values:
b= 2.5×106km ×√1−0.42
b= 2.5×106km ×√1−0.16
b= 2.5×106km ×√0.84
b= 2.5×106km ×0.916515
b≈2.2912875 ×106km
Step 4: Therefore, the distance between the planet and the Sun at its closest
approach is approximately 2.291 ×106km.
Question 25
Question
Suppose a planet has an elliptical orbit around the Sun such that its eccentricity
is 0.4. If the distance between the planet and the Sun at its closest approach
(perihelion) is 0.3 astronomical units (AU), determine the distance between the
planet and the Sun at its farthest point (aphelion).
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Solution
Step 1: Recall that the eccentricity of an elliptical orbit is defined as the ratio
of the distance between the foci of the ellipse to the length of the major axis.
Mathematically, the eccentricity (e) can be expressed as:
e=c
a
where cis half the distance between the foci of the ellipse, and ais half the
length of the major axis.
Step 2: We are given that the eccentricity of the planet’s orbit is 0.4, and
the distance between the planet and the Sun at perihelion is 0.3 AU. This
information allows us to set up an equation to solve for the distance between
the planet and the Sun at aphelion.
Step 3: Let the distance between the planet and the Sun at aphelion be x
AU. From the definition of eccentricity, we have:
0.4 = c
a=x−0.3
x+0.3
2
Step 4: Simplifying the equation, we get:
0.4 = 2(x−0.3)
x+ 0.3
Step 5: Cross multiplying gives:
0.4(x+ 0.3) = 2(x−0.3)
Step 6: Expanding both sides of the equation yields:
0.4x+ 0.12 = 2x−0.6
Step 7: Rearranging the terms gives:
0.12 + 0.6 = 2x−0.4x
0.72 = 1.6x
Step 8: Finally, solving for x, we find:
x=0.72
1.6≈0.45 AU
Step 9: Therefore, the distance between the planet and the Sun at its farthest
point (aphelion) is approximately 0.45 AU.
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